It's almost transparent. and . You should be able to see it from the above. Or you can argue as follows:

Yes there is: if n is a prime, obviously (n-1)! isn't divisible by n.

If n is not a prime and can be written as n = a * b with a != b then (n-1)! is divisible by n because it contains a and b.

If n = 4, (n-1)! isn't divisible by n, but if n = a * a with a being a prime number > 2, (n-1)! is divisible by n because we find a and 2a in (n-1)! (thanks to Juhana in comments).

Note that $(n-1)! = \color{blue}{(n-1)}(n-2)\cdots (2)(1) = \color{blue}{(n-1)}(n-2)!$ by plugging in $(n-2)! = (n-2)\cdots (2)(1)$ into the expression for $(n-1)!$.

So $$\frac{(n-1)!}{(n-2)!} = \frac{(n-1)(n-2)!}{(n-2)!} = n-1$$

$$\require{cancel}n!=\overbrace{1\times2\times3\times\dots\times(n-1)\times n}^n$$

$$\frac{n!}n=\frac{\overbrace{\color{green}{1\times2\times3\times\dots\times(n-1)}\times\color{red}{\cancel n}}^n}{\color{red}{\cancel n}}$$

$$(n-1)!=\overbrace{\color{green}{1\times2\times3\times\dots\times(n-1)}}^{n-1}$$

We also have:

$$\frac{n!}{p!}=\frac{\overbrace{\color{red}{\cancel{1\times2\times3\times\dots\times p}}\times\color{green}{(p+1)\times\dots\times(n-1)\times n}}^n}{\color{red}{\cancel{1\times2\times3\times\dots\times p}}}$$

$$=\color{green}{(p+1)\times\dots\times(n-1)\times n}$$